Proof by induction involving symmetric groups

Question

I am aware that i have posted this question before, but the comments i received did not really help silly old me. So i would appreciate it if someone can walk me through some parts of the problem. I have revised my attempt from the previous post, accounting for the comments i have received as i understood them.

Let $n>1$ be a natural number. the m-cycle
$\sigma=(a_1,a_2,a_3,\dots,a_m)$ denotes a permutation in $S_n$ where
$\sigma(a_1)=a_2,\sigma(a_2)=a_3,\dots, \text{ and }\sigma(a_m)=a_1$.

Prove by induction on $k \geq 1$ that if $k+i\equiv j (mod\text{ m})$, then $\sigma^k(a_i)=a_j$
whenever $1\leq i \leq m$ and $1\leq j \leq m$.

Pained Attempt :

Let $n>1$ be a natural number. the m-cycle $\sigma=(a_1,a_2,a_3,\dots,a_m)$ denotes a permutation in $S_n$ where
$\sigma(a_1)=a_2,\sigma(a_2)=a_3,\dots, \text{ and }\sigma(a_m)=a_1$.

[Base Step]

For the base case,$k=1$, assume that $1+i\equiv j (mod\text{ m})$ and prove that $\sigma^1(a_i)=a_j$ whenever $1\leq i \leq m$ and $1\leq j \leq m$.

Since $1+i\equiv j (mod\text{ m})$ then $j \equiv 1+i (mod\text{ m})$ and $m | j-(1+i)$ hence $j = mh+i+1$ for some $h \in \Bbb{Z}$.

[case:$1 \leq i \lt m$]

Since $j=mh+i+1$ then $a_j=a_{mh+i+1}$. By assumption $\sigma(a_{mh+i})=a_{mh+i+1}=a_j$.

I am not sure how to proceed further here..

I thought:
With $1 \leq i \lt m$ and mh+i+1 being constrained to lie between 1 and m given the definition of the m-cycle, this implies that h must be 0 (?).Therefore $\sigma(a_{m(0)+i})=a_{m(0)+i+1}=a_j$ and $\sigma(a_{i})=a_{i+1}=a_j$ as required (?)

…this feels iffy.

[case: $i=m$]

Since $j=mh+i+1$ then $a_j=a_{mh+m+1}$ and $a_j=a_{m(h+1)+1}$ . By assumption $\sigma(a_{m(h+1)})=a_{m(h+1)+1}=a_j$.

I am not sure how to proceed further either..

Any help would be much appreciated.

Answer 1

You are getting hung up on completely irrelevant details. Once you know $k$ and $i$, you know $j$. There's no need to delve into modular arithmetic and introduce new variables. Here's a complete proof of the base step:

Base case:

Let $k=1$ and let $1\leq i,j\leq m$ be such that $k+i\equiv j\pmod{m}$. We distinguish two cases:

1. If $1\leq i<m$ then $1<k+i\leq m$ and so $j=k+i=1+i$. Then $\sigma^1(a_i)=\sigma(a_i)=a_{i+1}=a_j$.
2. If $i=m$ then $k+i=m+1$ and so $j=1$. Then $\sigma^1(a_i)=\sigma(a_m)=a_1=a_j$.

Can you now prove the induction step?